Let $g(x)=x^3-16x$ and let $c$ be the number that satisfies the Mean Value Theorem for $g$ on the interval $[-4,2]$. What is $c$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $-3$ (Choice B) B $-2$ (Choice C) C $0$ (Choice D) D $1$
Answer: According to the Mean Value Theorem, there exists a number $c$ in the open interval $(-4,2)$ such that $g'(c)$ is equal to the average rate of change of $g$ over the interval: $g'(c)=\dfrac{g(2)-g(-4)}{(2)-(-4)}$ First, let's find that average rate of change: $\dfrac{g(2)-g(-4)}{(2)-(-4)}=\dfrac{-24-0}{6}={-4}$ Now, let's differentiate $g$ and find the $x$ -value for which $g'(x)={-4}$. $g'(x)=3x^2-16$ The solutions of $g'(x)=-4$ are $x=-2$ and $x=2$. Out of these, only $x=-2$ is within the interval $(-4,2)$. In conclusion, $c=-2$.